Rational expression word problem

[khavar]

"A boat travels 25 miles per hour in still water. It takes 3 and 1\3 hours to travel 40 miles up a river and then to return by the same route. What is the speed of the current in the river?"

Typed word-for-word from A-Plus Notes for Algebra (Algebra 2 and Pre-Calculus) by Rong Yang.

There is where I have given up:
40 divided by (25+r) = a fraction of 3 and 1\3 hours, represented by x
40 divided by (25-r) = 1-x
x(25+r)=(1-x)(25-r)
r=50x-25
Upon plugging r back into one of the above equations, it becomes evident that I have no idea what the ____ I am doing.

Thank you for your help.

 

[khavar]

I think I'm getting closer.
(1-x) seems to be wrong so I changed it to ((10\3)-x), which makes r equal to -15x+25. Although it gives me an answer of r=-10. Still wrong, but I think I am getting there.

Update 1:47 PM:

40\ (25+r) = (10\3)x
40\(25-r) = (10\3)(1-x)

2:03 PM.
Another dead-end. Solving for r gives me r^2+28r+675=0. Non-real solutions.

 

[Deleted member 4993]

"A boat travels 25 miles per hour in still water. It takes 3 and 1\3 hours to travel 40 miles up a river and then to return by the same route. What is the speed of the current in the river?"

Typed word-for-word from A-Plus Notes for Algebra (Algebra 2 and Pre-Calculus) by Rong Yang.

There is where I have given up:
40 divided by (25+r) = a fraction of 3 and 1\3 hours, represented by x
40 divided by (25-r) = 1-x
x(25+r)=(1-x)(25-r)
r=50x-25
Upon plugging r back into one of the above equations, it becomes evident that I have no idea what the ____ I am doing.

Thank you for your help.

Let the speed of the current = C

Then

speed of the boat with the current = 25 + C

speed of the boat against the current = 25 - C

To travel 40 miles with the current time requird would be = 40/(25+C)

To travel 40 miles with the current time requird would be = 40/(25- C)

Then

40/(25+C) + 40/(25-C) = 10/3

50/(625-C²) = 1/12

625 - C² = 600

Solve for C from and then check for correctness.

 

[khavar]

Let the speed of the current = C

Then

speed of the boat with the current = 25 + C

speed of the boat against the current = 25 - C

To travel 40 miles with the current time requird would be = 40/(25+C)

To travel 40 miles with the current time requird would be = 40/(25- C)

Then

40/(25+C) + 40/(25-C) = 10/3

50/(625-C²) = 1/12

625 - C² = 600

Solve for C from and then check for correctness.

Thank goodness. Thank you, Subhotosh. The current equals 5 miles per hour.
I had been working on that problem since 11:30 Am. I love math, I wish I were better at it though.

 

[mmm4444bot]

It takes 3 and 1\3 hours to travel 40 miles up a river and then to return by the same route.

What is the speed of the current in the river?"

40 divided by (25+r) = a fraction of 3 and 1\3 hours, represented by x

40 divided by (25-r) = 1-x

evident that I have no idea what the ____ I am doing.

I disagree. You developed expressions 40/(25+r) and 40/(25-r) as the times for each leg of the trip. That's more than what many people posting this exercise come up with.

You simply failed to realize that your two time expressions add up to the total trip time (read blue statement above). Learning math is a process of making mistakes, understanding them, and moving on. Now you're in a better position to not make this mistake on the next word problem involving two things and their total.

Here's another clue: the red part of the question above. Note that they are asking for ONE quantity. This is not an exercise where you set up an equation with two unknowns to solve (x and r).


You used the symbol r to represent the current's rate. It's good practice to always state up-front symbol definitions; not only does this communicate to others what you're thinking, but it also gives you something to look at when you need to refocus on what you're trying to accomplish.

"Let r = current's rate"


By the way, if you were thinking -- that by assigning x to represent some fraction of the trip -- you could let 1-x represent the other fraction (using "1" to represent "the whole"), that does not help because the exercise doesn't provide information about how the two fractions relate to one another. In other words, all we would know about their relationship is the following, which is not very helpful.

x + (1 - x) = 1

1 = 1

Cheers :cool:

 

[khavar]

I disagree. You developed expressions 40/(25+r) and 40/(25-r) as the times for each leg of the trip. That's more than what many people posting this exercise come up with.

You simply failed to realize that your two time expressions add up to the total trip time (read blue statement above). Learning math is a process of making mistakes, understanding them, and moving on. Now you're in a better position to not make this mistake on the next word problem involving two things and their total.

Here's another clue: the red part of the question above. Note that they are asking for ONE quantity. This is not an exercise where you set up an equation with two unknowns to solve (x and r).


You used the symbol r to represent the current's rate. It's good practice to always state up-front symbol definitions; not only does this communicate to others what you're thinking, but it also gives you something to look at when you need to refocus on what you're trying to accomplish.

"Let r = current's rate"


By the way, if you were thinking -- that by assigning x to represent some fraction of the trip -- you could let 1-x represent the other fraction (using "1" to represent "the whole"), that does not help because the exercise doesn't provide information about how the two fractions relate to one another. In other words, all we would know about their relationship is the following, which is not very helpful.

x + (1 - x) = 1

1 = 1

Cheers :cool:

Thank you, M4bot. Later in the evening yesterday, I made myself some dinner and sat patiently with another problem that was giving me some trouble. I had a profound sense of patience and working the problem made me happy.
It has taken me years but I am beginning to understand to not be so hard on myself and to just keep going with the problem patiently.